A={bmatrix} x+y & y \\ x & x-y {bmatrix}, B={bmatrix} 3 \\ -2 {bmatrix} and C={bmatrix} 4 \\ -2 {bmatrix} If AB = C, then what is A² equal to?

Correct Answer is : [−4−8812]{bmatrix} -4 & -8 \\ 8 & 12 {bmatrix}[−48−812]

{bmatrix} 4 & 8 \\ -4 & -16 {bmatrix}

{bmatrix} 4 & -4 \\ 8 & -16 {bmatrix}

{bmatrix} -4 & -8 \\ 4 & 12 {bmatrix}

{bmatrix} -4 & -8 \\ 8 & 12 {bmatrix}

Correct Answer is : [−4−8812]{bmatrix} -4 & -8 \\ 8 & 12 {bmatrix}[−48−812]

Test Index

UPSC NDA I 2017 Mathematics Solved Paper

Question : 35 of 120

Marks: +1, -0

A=\begin{bmatrix} x+y & y \\ x & x-y \end{bmatrix}, B=\begin{bmatrix} 3 \\ -2 \end{bmatrix}

C=\begin{bmatrix} 4 \\ -2 \end{bmatrix}

AB = C

A^{2}

equal to?

Solution:

Concept:

Use matrix multiplication to set up equations from

AB = C

. Solve for

x

and

y

, then compute

A^2

Explanation:

Given

A = \begin{bmatrix} x+y & y \\ x & x-y \end{bmatrix}

B = \begin{bmatrix} 3 \\ -2 \end{bmatrix}

C = \begin{bmatrix} 4 \\ -2 \end{bmatrix}

We have

AB = C

Multiply:

\begin{bmatrix} x+y & y \\ x & x-y \end{bmatrix} \begin{bmatrix} 3 \\ -2 \end{bmatrix} = \begin{bmatrix} 3(x+y) + y(-2) \\ 3x + (x-y)(-2) \end{bmatrix} = \begin{bmatrix} 3x+y \\ x+2y \end{bmatrix}

Set this equal to

C

\begin{bmatrix} 3x+y \\ x+2y \end{bmatrix} = \begin{bmatrix} 4 \\ -2 \end{bmatrix}

This yields equations:

3x+y=4

and

x+2y=-2

Solve: From first,

y=4-3x

. Substitute into second:

x+2(4-3x)=-2

gives

x+8-6x=-2

, so

-5x=-10

x=2

Then

y=4-6=-2

Now

A = \begin{bmatrix} 2+(-2) & -2 \\ 2 & 2-(-2) \end{bmatrix} = \begin{bmatrix} 0 & -2 \\ 2 & 4 \end{bmatrix}

Compute

A^2 = A \cdot A

\begin{bmatrix} 0 & -2 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} 0 & -2 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} 0\cdot0+(-2)\cdot2 & 0\cdot(-2)+(-2)\cdot4 \\ 2\cdot0+4\cdot2 & 2\cdot(-2)+4\cdot4 \end{bmatrix} = \begin{bmatrix} -4 & -8 \\ 8 & 12 \end{bmatrix}

Answer:

A^2 = \begin{bmatrix} -4 & -8 \\ 8 & 12 \end{bmatrix}

, which corresponds to option D.

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